\( 1<\quad \) Use any method to evaluate the integral. \( \int \frac{22 d x}{\left(121 x^{2}+1\right)^{2}} \)

To evaluate the integral \[ \int \frac{22 \, dx}{(121 x^{2}+1)^{2}}, \] we can use a trigonometric substitution. Notice that the expression \(121 x^2 + 1\) resembles the identity for \( \sec^2(\theta) \). We can let \[ x = \frac{1}{11} \tan(\theta), \] then \[ dx = \frac{1}{11} \sec^2(\theta) d\theta. \] This gives us \[ 121x^2 + 1 = 121\left(\frac{1}{11}\tan(\theta)\right)^2 + 1 = \tan^2(\theta) + 1 = \sec^2(\theta). \] Hence, \[ (121x^2 + 1)^2 = (\sec^2(\theta))^2 = \sec^4(\theta). \] Now substituting in the integral, we get \[ \int \frac{22 \cdot \frac{1}{11} \sec^2(\theta) \, d\theta}{\sec^4(\theta)} = \int \frac{2 \sec^2(\theta) \, d\theta}{\sec^4(\theta)} = \int 2 \cos^2(\theta) \, d\theta = 2 \int \cos^2(\theta) \, d\theta. \] Using the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \): \[ 2 \int \cos^2(\theta) \, d\theta = 2 \cdot \frac{1}{2} \int (1 + \cos(2\theta)) \, d\theta = \int (1 + \cos(2\theta)) \, d\theta = \theta + \frac{1}{2} \sin(2\theta) + C. \] We now need to revert back to \(x\): \[ \theta = \tan^{-1}(11x) \quad \text{and} \quad \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \cdot \frac{11x}{\sqrt{121x^2 + 1}} \cdot \frac{1}{\sqrt{121x^2 + 1}} = \frac{22x}{121x^2 + 1}. \] Thus, the integral evaluates to \[ \tan^{-1}(11x) + \frac{11x}{121x^2 + 1} + C. \] Final answer: \[ \int \frac{22 \, dx}{(121 x^{2}+1)^{2}} = \tan^{-1}(11x) + \frac{11x}{121x^2 + 1} + C. \]